Question

If ${\int }_0^{2\pi }\left|xsinx\right|dx=k\pi ,$ then the values of $k$ is equal to

• A. 4

Answer 1

The correct option is A 4
Let $I={\int }_0^{2\pi }\left|xsinx\right|dx$
$={\int }_0^{2\pi }x\left|sinx\right|dx[\because \left|x\right|=xforx>0]$
$I={\int }_0^{2\pi }(2\pi -x)\left|sinx\right|dx[$Using Note $2$ below$]$
$2I={\int }_0^{2\pi }2\pi \left|sinx\right|dx$
$I=\pi {\int }_0^{2\pi }\left|sinx\right|dx$
$=2\pi {\int }_0^{\pi }\left|sinx\right|dx[$using note $\left(1\right)$ below$]$
$=4\pi {\int }_0^{\pi /2}\left|sinx\right|dx=4\pi [-cosx{]}_0^{\pi /2}$
$I=4\pi$
So, ${\int }_0^{2\pi }\left|xsinx\right|dx=k\pi$
$\Rightarrow 4\pi =k\pi \Rightarrow k=4$

Note: $\left(1\right){\int }_0^{2a}f\left(x\right)dx=\{\begin{array}{cc}2{\int }_0^{a}f\left(x\right)dx,& iff(2a-x)=f\left(x\right)\\ 0,& iff(2a-x)=-f\left(x\right)\end{array}$
$\left(2\right){\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$